Variance of Processing Time for Different Policies

1
Variance of Processing Time for Different Policies

• Assume that there are n processes on the system
  and process 1 lt k lt n has processing time equal
  to TBk2k.
• Compute average response (turnaround) Tt time for
  these processes assuming
• 1. arrival in ascending order of process numbers
  and FCFS schedule,
• 2. arrival in descending order of process numbers
  and FCFS schedule,
• 3. arrival in ascending order of process numbers
  and RR schedule with time slice equal to 1,

2
Variance of Processing Time for Different Policies

• 4. arrival in descending order of process numbers
  and RR schedule with time slice equal to 1.
• Compare the difference between Tts of the same
  scheduling policy for different orders of
  processes.
• Exercise recalculate variance of processing time
  for TBk k.

3
Solution

• Let tw(k) denote the waiting time of process k,
  Tw average waiting time. Then

and
Tw
where Tt denotes the average turnaround time
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Solution of Case 1
5
Solution of Case 2
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Solution of Case 3

• Let tp(k) denote the processing time for process
  k, where 2 lt k lt n

tp(k) tp(k-1) 1 (n - k 1)2k-1
Reason this process in one time unit completes
the round which was last for process k-1, and
then makes 2k-1 additional rounds in RR schedule
compared to process k-1. Each of this additional
rounds has n-k1 processes in it. We also have
tp(1) n 1 because the first round for process
1 takes n units and the last round just one unit.
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Solution of Case 3 continued
so,
and finally,
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Solution of Case 4

• In this case tp(1)2n and for k gt 1

Reason all 2k-1 rounds which process k executes
after process k-1 finished have all n-k1
remaining processes executed before process k
finishes. Hence,
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Solution of Case 4 continued
10
Variance of Processing Time for tBkk (Exercise)
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Solution of Case 3 for the Exercise
Let tp(k) denote the processing time of k-th
process, then for kgt1 tp(k) tp(k-1)
n-k2 because process k makes one round in RR
schedule with n-(k-1) processes left and then one
round for the last work unit and clearly tp(1)
1, so
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Solution of Case 4 for the Exercise
In this case tp(1)n and for kgt1 tp(k)
tp(k-1)n-k1 because the round for process k
has n-k1 processes executed before and with it,
so